Question

我有一个包含三个不同字段（Date，Time，Value）的数据集，如下所示：

2014-01-24 23:32:01.874 45
2014-01-24 23:32:02.198 71
2014-01-24 23:32:02.302 94
2014-01-24 23:32:02.439 48
2014-01-24 23:32:02.574 82
2014-01-24 23:32:02.724 51
2014-01-24 23:32:02.913 15
2014-01-24 23:32:02.964 77
2014-01-24 23:32:02.989 49
2014-01-24 23:32:03.017 42
2014-01-24 23:32:03.025 1
2014-01-24 23:32:03.085 67
2014-01-24 23:32:03.136 53
2014-01-24 23:32:03.200 46
2014-01-24 23:32:03.240 72
2014-01-24 23:32:03.257 0
2014-01-24 23:32:03.296 36

如何编写Perl脚本来计算每5分钟的平均值？

期望的输出：

Time  Average Value
23:30 20
23:35 35
23:40 15

Answer 1

编写Perl代码以逐行读取数据。

维护一个计数器变量，以5分钟为增量计算时间，以及一个包含您正在读取的数据行的数组。

当数据行的时间超过当前计数器变量值时，使用数组中的数据计算平均值，递增计数器并清除阵列。将新行添加到已清除的数组中，然后控制该过程。

Answer 2

我将数据更改为间隔为5分钟：

2014-01-24 23:12:01.874 45
2014-01-24 23:12:02.198 71
2014-01-24 23:22:02.302 94
2014-01-24 23:22:02.439 48
2014-01-24 23:22:02.574 82
2014-01-24 23:32:02.724 51
2014-01-24 23:32:02.913 15
2014-01-24 23:32:02.964 77
2014-01-24 23:42:02.989 49
2014-01-24 23:42:03.017 42
2014-01-24 23:42:03.025 1
2014-01-24 23:52:03.085 67
2014-01-24 23:52:03.136 53
2014-01-24 23:52:03.200 46
2014-01-24 23:52:03.240 72
2014-01-24 23:52:03.257 0
2014-01-24 23:52:03.296 36

剧本：

#!/usr/bin/perl -n

BEGIN { use Date::Parse; }

push @r, [ $&, str2time($1).'.'.$2, $3 ] if /(.*?)\.(\d+) (\d+)\s*$/;

END {

  $ref = undef;
  $avg = 0.0;
  $cnt = 0;
  for(sort { $a->[1] <=> $b->[1]} @r) {
      unless ($ref) {
        $sum = $_->[2];
        $cnt = 1;
        $ref = $_->[1];
      } else {
        if ($_->[1] - $ref > 60.0*5) {
          printf "avg: %5.2f\n", $sum/$cnt;
          $ref = $_->[1];
          $sum = $_->[2];
          $cnt = 1;
        } else {
          $sum += $_->[2];
          $cnt++;
        }
      }
      printf "%s", $_->[0];
  }
  printf "avg: %5.2f\n", $sum/$cnt;
}

输出：

2014-01-24 23:12:01.874 45
2014-01-24 23:12:02.198 71
avg: 58.00
2014-01-24 23:22:02.302 94
2014-01-24 23:22:02.439 48
2014-01-24 23:22:02.574 82
avg: 74.67
2014-01-24 23:32:02.724 51
2014-01-24 23:32:02.913 15
2014-01-24 23:32:02.964 77
avg: 47.67
2014-01-24 23:42:02.989 49
2014-01-24 23:42:03.017 42
2014-01-24 23:42:03.025 1
avg: 30.67
2014-01-24 23:52:03.085 67
2014-01-24 23:52:03.136 53
2014-01-24 23:52:03.200 46
2014-01-24 23:52:03.240 72
2014-01-24 23:52:03.257 0
2014-01-24 23:52:03.296 36
avg: 45.67

Answer 3

您可以将时间戳标准化为五分钟，通过转换为秒数除以300，取整数并乘以300得分。

采用这种方法（并使用Perl的标准Time :: Piece模块进行解析和格式化）得出如下内容：

#!/usr/bin/perl

use strict;
use warnings;
use 5.010;

use Time::Piece;

my $fmt = '%Y-%m-%d %H:%M:%S';

my %data;
while (<DATA>) {
  chomp;
  my ($date, $time, undef, $count) = split /[ \.]/;

  my $dt = Time::Piece->strptime("$date $time", $fmt);
  push @{$data{five_min($dt->epoch)}}, $count;
}

foreach my $period (sort keys %data) {
  my $total;
  $total += $_ for @{$data{$period}};
  my $avg = sprintf '%.2f', $total / @{$data{$period}};
  say localtime($period)->strftime($fmt), ' ', $avg;
}

sub five_min {
  my $epoch = shift;
  my $secs  = 5 * 60;
  return $secs * int($epoch / $secs);
}

__DATA__
2014-01-24 23:12:01.874 45
2014-01-24 23:12:02.198 71
2014-01-24 23:22:02.302 94
2014-01-24 23:22:02.439 48
2014-01-24 23:22:02.574 82
2014-01-24 23:32:02.724 51
2014-01-24 23:32:02.913 15
2014-01-24 23:32:02.964 77
2014-01-24 23:42:02.989 49
2014-01-24 23:42:03.017 42
2014-01-24 23:42:03.025 1
2014-01-24 23:52:03.085 67
2014-01-24 23:52:03.136 53
2014-01-24 23:52:03.200 46
2014-01-24 23:52:03.240 72
2014-01-24 23:52:03.257 0
2014-01-24 23:52:03.296 36

输出结果为：

$ ./avgtime 
2014-01-24 23:10:00 58.00
2014-01-24 23:20:00 74.67
2014-01-24 23:30:00 47.67
2014-01-24 23:40:00 30.67
2014-01-24 23:50:00 45.67

计算Perl中5分钟的平均值

3 个答案: